How do you find three consecutive even integers have a sum of -36?

2 Answers
Mar 10, 2018

Answer:

See a solution process below:

Explanation:

Let's call the first even integer: #n#

Then the next two even integers would be: #n + 2#; #n + 4#

Sum means to add so we can write the following equation and solve for #n#:

#n + (n + 2) + (n + 4) = -36#

#n + n + 2 + n + 4 = -36#

#n + n + n + 2 + 4 = -36#

#1n + 1n + 1n + 6 = -36#

#(1 + 1 + 1)n + 6 = -36#

#3n + 6 = -36#

#3n + 6 - color(red)(6) = -36 - color(red)(6)#

#3n + 0 = -42#

#3n = -42#

#(3n)/color(red)(3) = -42/color(red)(3)#

#(color(red)(cancel(color(black)(3)))n)/cancel(color(red)(3)) = -14#

#n = -14#

Then:

#n + 2 = -14 + 2 = -12#

#n + 4 = -14 + 4 = -10#

#(-14) + (-12) + (-10) = -36

The three consecutive even integers adding up to -36 are:

-14, -12, -10

Mar 10, 2018

Answer:

#-14,-12,-10.#

Explanation:

Take three consecutive even integers:
#color(red)((2n-2),2n,(2n+2))#
#:.# sum #=(2n-2)+2n+(2n-2)=-36#
#=>2n-cancel(2)+2n+2n+cancel(2)=-36#
#=>6n=-36=>n=-6#
So,the required integers:
#color(red)(2n-2)=2(-6)-2=-14#
#color(red)(2n)=2(-6)=-12#
#color(red)(2n+2)=2(-6)+2=-10#